Is there a combination of takeWhile,dropWhile in Scala? - string

In Scala, I want to split a string at a specific character like so:
scala> val s = "abba.aadd"
s: String = abba.aadd
scala> val (beforeDot,afterDot) = (s takeWhile (_!='.'), s dropWhile (_!='.'))
beforeDot: String = abba
afterDot: String = .aadd
This solution is slightly inefficient (maybe not asymptotically), but I have the feeling something like this might exist in the standard library already. Any ideas?

There is a span method:
scala> val (beforeDot, afterDot) = s.span{ _ != '.' }
beforeDot: String = abba
afterDot: String = .aadd
From the Scala documentation:
c span p is equivalent to (but possibly more efficient than) (c takeWhile p, c dropWhile p), provided the evaluation of the predicate p does not cause any side-effects.

You can use splitAt for what you want:
val s = "abba.aadd"
val (before, after) = s.splitAt(s.indexOf('.'))
Output:
before: String = abba
after: String = .aadd

Related

Merge two strings in kotlin

I have two strings
val a = "abc"
val b = "xyz"
I want to merge it and need output like below
axbycz
I added both strings to arraylist and then flatmap it
val c = listOf(a, b)
val d = c.flatMap {
it.toList()
}
but not getting the desired result
Use the zip function. It creates a list of pairs with "adjacent" letters. You can then use joinToString with a transformer to create your final result.
a.zip(b) // Returns the list [(a, x), (b, y), (c, z)]
.joinToString("") { (a, b) -> "$a$b" } // Joins the list back to a string with no separator
You can always use a simple loop, assuming both strings have the same size. That way You only allocate a StringBuilder and counter variable, without any lists, arrays or pairs:
val a = "abc"
val b = "xyz"
val sb = StringBuilder()
for(i in 0 until a.length){
sb.append(a[i]).append(b[i])
}
val d = sb.toString()
marstran's answer is really concise and Pawels answer is really fast. Using buildString you can have to best of both worlds:
buildString {
a.zip(b).forEach { (a, b) ->
append(a).append(b)
}
}
buildString creates a StringBuilder and offers it as receiver in the lambda. It returns the built string.
Try it out here: Kotlin Playground. Thanks to Pawel for creating the original benchmark.

scala - string parsing without Regex

I have various types of strings like the following:
sales_data_type
saledatatypes
sales_data.new.metric1
sales_data.type.other.metric2
sales_data.type3.metric3
I'm trying to parse them to get a substring with a word before and after the last dot. For example: new.metric1, other.metric2, type3.metric3. If a word doesn't contain dots, it has to be returned as is: sales_data_type, saledatatypes.
With a Regex it may be done this way:
val infoStr = "sales_data.type.other.metric2"
val pattern = ".*?([^.]+\\.)?([^.]+)$"
println(infoStr.replaceAll(pattern, "$1$2"))
// prints other.metric2
// for saledatatypes just prints nullsaledatatypes ??? but seems to work
I want to find a way to achieve this with Scala, without using Regex in order to expand my understanding of Scala features. Will be grateful for any ideas.
One-liner:
dataStr.split('.').takeRight(2).mkString(".")
takeRight(2) will take the last 2 if there are 2 to take, else it will take the last, and only, 1. mkString(".") will re-insert the dot only if there are 2 elements for the dot to go between, else it will leave the string unaltered.
Here's one with lots of scala features for you.
val string = "head.middle.last"
val split = string.split('.') // Array(head, middle, last)
val result = split.toSeq match {
case Seq(word) ⇒ word
case _ :+ before :+ after ⇒ s"$before.$after"
}
println(result) // middle.last
First we split the string on your . and get individual parts.
Then we pattern match those parts, first to check if there is only one (in which case we just return it), and second to grab the last two elements in the seq.
Finally we put a . back in between those last two using string interpolation.
One way of doing it:
val value = "sales_data.type.other.metric2"
val elems = value.split("\\.").toList
elems match {
case _:+beforeLast:+last => s"${beforeLast}.${last}"
case _ => throw new NoSuchElementException
}
for(s<-strs) yield {val s1 = s.split('.');
if(s1.size>=2)s1.takeRight(2).mkString(".") else s }
or
for(s<-strs) yield { val s1 = s.split('.');
if(s1.size>=2)s1.init.last+'.'+s1.last else s }
In Scala REPL:
scala> val strs =
Vector("sales_data_type","saledatatypes","sales_data.new.metric1","sales_data.type.other.metric2","sales_d
ata.type3.metric3")
strs: scala.collection.immutable.Vector[String] = Vector(sales_data_type, saledatatypes, sales_data.new.metric1, sales_data.
type.other.metric2, sales_data.type3.metric3)
scala> for(s<-strs) yield { val s1 = s.split('.');if(s1.size>=2)s1.takeRight(2).mkString(".") else s }
res62: scala.collection.immutable.Vector[String] = Vector(sales_data_type, saledatatypes, new.metric1, other.metric2, type3.
metric3)
scala> for(s<-strs) yield { val s1 = s.split('.');if(s1.size>=2)s1.init.last+'.'+s1.last else s }
res60: scala.collection.immutable.Vector[String] = Vector(sales_data_type, saledatatypes, new.metric1, other.metric2, type3.
metric3)
Use scala match and do like this
def getFormattedStr(str:String):String={
str.contains(".") match{
case true=>{
val arr=str.split("\\.")
val len=arr.length
len match{
case 1=>str
case _=>arr(len-2)+"."+arr(len-1)
}
}
case _=>str
}
}

Scala - modify strings in a list based on their number of occurences

Another Scala newbie question since I am not getting how to achieve this in a functional way (mostly coming from a scripting language background):
I have a list of strings:
val food-list = List("banana-name", "orange-name", "orange-num", "orange-name", "orange-num", "grape-name")
and where they are duplicated, I'd like to add an incrementing number into the string and get that in a list similar to the input list, like so:
List("banana-name", "orange1-name", "orange1-num", "orange2-name", "orange2-num", "grape-name")
I've grouped them up to get counts for them with:
val freqs = list.groupBy(identity).mapValues(v => List.range(1, v.length + 1))
Which gives me:
Map(orange-num -> List(1, 2), banana-name -> List(1), grape-name -> List(1), orange-name -> List(1, 2))
The order of the list is important (it should be in the original order of food-list) so I know it's problematic for me to use a Map at this point. The closest I feel I have gotten to a solution is:
food-list.map{l =>
if (freqs(l).length > 1){
freqs(l).map(n =>
l.split("-")(0) + n.toString + "-" + l.split("-")(1))
} else {
l
}
}
This of course gives me a wonky output since I am mapping the list of frequencies from the words value in freqs
List(banana-name, List(orange1-name, orange2-name), List(orange1-num, orange2-num), List(orange1-name, orange2-name), List(orange1-num, orange2-num), grape-name)
How is this done in a Scala fp way without resorting to clumsy for loops and counters?
If the indices are important, sometimes it's best to keep track of them explicitly using zipWithIndex (very similar to Python's enumerate):
food-list.zipWithIndex.groupBy(_._1).values.toList.flatMap{
//if only one entry in this group, don't change the values
//x is actually a tuple, could write case (str, idx) :: Nil => (str, idx) :: Nil
case x :: Nil => x :: Nil
//case where there are duplicate strings
case xs => xs.zipWithIndex.map {
//idx is index in the original list, n is index in the new list i.e. count
case ((str, idx), n) =>
//destructuring assignment, like python's (fruit, suffix) = ...
val Array(fruit, suffix) = str.split("-")
//string interpolation, returning a tuple
(s"$fruit${n+1}-$suffix", idx)
}
//We now have our list of (string, index) pairs;
//sort them and map to a list of just strings
}.sortBy(_._2).map(_._1)
Efficient and simple:
val food = List("banana-name", "orange-name", "orange-num",
"orange-name", "orange-num", "grape-name")
def replaceName(s: String, n: Int) = {
val tokens = s.split("-")
tokens(0) + n + "-" + tokens(1)
}
val indicesMap = scala.collection.mutable.HashMap.empty[String, Int]
val res = food.map { name =>
{
val n = indicesMap.getOrElse(name, 1)
indicesMap += (name -> (n + 1))
replaceName(name, n)
}
}
Here is an attempt to provide what you expected with foldLeft:
foodList.foldLeft((List[String](), Map[String, Int]()))//initial value
((a/*accumulator, list, map*/, v/*value from the list*/)=>
if (a._2.isDefinedAt(v))//already seen
(s"$v+${a._2(v)}" :: a._1, a._2.updated(v, a._2(v) + 1))
else
(v::a._1, a._2.updated(v, 1)))
._1/*select the list*/.reverse/*because we created in the opposite order*/

Unable to access String value

Why unable to access the String value ?
I would expect the s1 to be "a" but instead its Ljava.lang.String;#d70d7a ?
val it = Iterator("(a,((a,b),1.0))") //> it : Iterator[String] = non-empty iterator
val s1 = it.next.replace("(" , "").replace(")" , "").split(",").toString.split(",")
//> s1 : Array[String] = Array([Ljava.lang.String;#d70d7a)
println("s1 is "+s1(0)) //> s1 is [Ljava.lang.String;#d70d7a
Let's go command-by-command:
val it = Iterator("(a,((a,b),1.0))")
// here we got iterator on one String
val s1 = it.next // "(a,((a,b),1.0))"
.replace("(" , "") // "a,a,b),1.0))"
.replace(")" , "") // "a,a,b,1.0"
.split(",") // multiple lines in array: "a", "a","b","1.0"
.toString // Array[String].toString returns what you got: Ljava.lang.String;#d70d7a
.split(",") // one String (because there's no "," signs)
Maybe you should run toList before toString, because toString is defined in a way you expect it to be defined in this implementation of List:
val s1 = ...
.split(",")
.toList
.toString
...
Maybe you should look at Java: split() returns [Ljava.lang.String;#186d4c1], why? for clarification.
.split(",") makes Array and .toString doesn't work on Array and after that you are again spliting by .split(",")
which i don't think helpful.
and you can also use replaceAll in place of multiple replace
scala> val it = Iterator("(a,((a,b),1.0))")
it: Iterator[String] = non-empty iterator
scala> val s1 = it.next.replaceAll("[()]" , "").split(",")
s1: Array[String] = Array(a, a, b, 1.0)
scala> println("s1 is "+s1(0))
s1 is a

HowTo get a Map from a csv string

I'm fairly new to Scala, but I'm doing my exercises now.
I have a string like "A>Augsburg;B>Berlin". What I want at the end is a map
val mymap = Map("A"->"Augsburg", "B"->"Berlin")
What I did is:
val st = locations.split(";").map(dynamicListExtract _)
with the function
private def dynamicListExtract(input: String) = {
if (input contains ">") {
val split = input split ">"
Some(split(0), split(1)) // return key , value
} else {
None
}
}
Now I have an Array[Option[(String, String)
How do I elegantly convert this into a Map[String, String]
Can anybody help?
Thanks
Just change your map call to flatMap:
scala> sPairs.split(";").flatMap(dynamicListExtract _)
res1: Array[(java.lang.String, java.lang.String)] = Array((A,Augsburg), (B,Berlin))
scala> Map(sPairs.split(";").flatMap(dynamicListExtract _): _*)
res2: scala.collection.immutable.Map[java.lang.String,java.lang.String] = Map((A,Augsburg), (B,Berlin))
For comparison:
scala> Map("A" -> "Augsburg", "B" -> "Berlin")
res3: scala.collection.immutable.Map[java.lang.String,java.lang.String] = Map((A,Augsburg), (B,Berlin))
In 2.8, you can do this:
val locations = "A>Augsburg;B>Berlin"
val result = locations.split(";").map(_ split ">") collect { case Array(k, v) => (k, v) } toMap
collect is like map but also filters values that aren't defined in the partial function. toMap will create a Map from a Traversable as long as it's a Traversable[(K, V)].
It's also worth seeing Randall's solution in for-comprehension form, which might be clearer, or at least give you a better idea of what flatMap is doing.
Map.empty ++ (for(possiblePair<-sPairs.split(";"); pair<-dynamicListExtract(possiblePair)) yield pair)
A simple solution (not handling error cases):
val str = "A>Aus;B>Ber"
var map = Map[String,String]()
str.split(";").map(_.split(">")).foreach(a=>map += a(0) -> a(1))
but Ben Lings' is better.
val str= "A>Augsburg;B>Berlin"
Map(str.split(";").map(_ split ">").map(s => (s(0),s(1))):_*)
--or--
str.split(";").map(_ split ">").foldLeft(Map[String,String]())((m,s) => m + (s(0) -> s(1)))

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